Problem: Boboniu gives you

  * r red balls, 
  * g green balls, 
  * b blue balls, 
  * w white balls. 



He allows you to do the following operation as many times as you want: 

  * Pick a red ball, a green ball, and a blue ball and then change their color to white. 



You should answer if it's possible to arrange all the balls into a palindrome after several (possibly zero) number of described operations. 

Input

The first line contains one integer T (1≤ T≤ 100) denoting the number of test cases.

For each of the next T cases, the first line contains four integers r, g, b and w (0≤ r,g,b,w≤ 10^9).

Output

For each test case, print "Yes" if it's possible to arrange all the balls into a palindrome after doing several (possibly zero) number of described operations. Otherwise, print "No".

Example

Input


4
0 1 1 1
8 1 9 3
0 0 0 0
1000000000 1000000000 1000000000 1000000000


Output


No
Yes
Yes
Yes

Note

In the first test case, you're not able to do any operation and you can never arrange three balls of distinct colors into a palindrome.

In the second test case, after doing one operation, changing (8,1,9,3) to (7,0,8,6), one of those possible palindromes may be "rrrwwwbbbbrbbbbwwwrrr".

A palindrome is a word, phrase, or sequence that reads the same backwards as forwards. For example, "rggbwbggr", "b", "gg" are palindromes while "rgbb", "gbbgr" are not. Notice that an empty word, phrase, or sequence is palindrome.

Solution (in PYTHON):

Solution:
from __future__ import division, print_function


def main():
    # Template 1.0
    import sys, re, math
    from collections import deque, defaultdict, Counter, OrderedDict
    from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians, floor
    from heapq import heappush, heappop, heapify, nlargest, nsmallest
    def STR(): return list(input())

    def INT(): return int(input())

    def MAP(): return map(int, input().split())

    def LIST(): return list(map(int, input().split()))

    def list2d(a, b, c): return [[c] * b for i in range(a)]

    def sortListWithIndex(listOfTuples, idx):   return (sorted(listOfTuples, key=lambda x: x[idx]))

    def sortDictWithVal(passedDic):
        temp = sorted(passedDic.items(), key=lambda kv: (kv[1], kv[0]))
        toret = {}
        for tup in temp:
            toret[tup[0]] = tup[1]
        return toret

    def sortDictWithKey(passedDic):
        return dict(OrderedDict(sorted(passedDic.items())))

    INF = float('inf')
    mod = 10 ** 9 + 7

    t = INT()

    while(t!=0):

        r,g,b,w = MAP()

        top = (r%2==1)+(g%2==1)+(b%2==1)+(w%2==1)

        if(top<=1):
            print("Yes")
        else:
            if(r>=1 and b>=1 and g>=1):
                r-=1
                g-=1
                b-=1
                w+=3
                top = (r % 2 == 1) + (g % 2 == 1) + (b % 2 == 1) + (w % 2 == 1)
                if (top <= 1):
                    print("Yes")
                else:
                    print("No")
            else:
                print("No")
        t-=1




######## Python 2 and 3 footer by Pajenegod and c1729

# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.

# So on cf, use PyPy2 for best string performance.

py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip

    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192


class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0, 2), super(FastIO, self).write(s))[0])
        return s

    def read(self):
        while self._fill(): pass
        return super(FastIO, self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill();
            self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s: self.buffer.write(s.encode('ascii'))
            self.read = lambda: self.buffer.read().decode('ascii')
            self.readline = lambda: self.buffer.readline().decode('ascii')


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')

# Cout implemented in Python
import sys


class ostream:
    def __lshift__(self, a):
        sys.stdout.write(str(a))
        return self


cout = ostream()
endl = '\n'


# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero=0):
    conv = ord if py2 else lambda x: x
    A = [];
    numb = zero;
    sign = 1;
    i = 0;
    s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'0'[0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-'[0]:
                sign = -1
            elif s[i] != b'\r'[0]:
                A.append(sign * numb)
                numb = zero;
                sign = 1
            i += 1
    except:
        pass
    if s and s[-1] >= b'0'[0]:
        A.append(sign * numb)
    return A


if __name__ == "__main__":
    main()